CBSE Class 12 Chemistry Electrochemistry Worksheet Set 05

Subjective Questions

Short Answer Type Questions-I

Question. Calculate the degree of dissociation \( (\alpha) \) of acetic acid if its molar conductivity \( (\Lambda_m) \) is \( 39.05 \text{ S cm}^2 \text{ mol}^{-1} \). Given \( \Lambda^\circ_{(H^+)} = 349.6 \text{ S cm}^2 \text{ mol}^{-1} \) and \( \Lambda^\circ_{(CH_3COO^-)} = 40.9 \text{ S cm}^2 \text{ mol}^{-1} \).
Answer: \( \Lambda^\circ_{CH_3COOH} = \Lambda^\circ_{CH_3COO^-} + \Lambda^\circ_{H^+} \)
\( = 40.9 + 349.6 = 390.5 \text{ S cm}^2/\text{mol} \)
Now, \( \alpha = \Lambda_m / \Lambda^\circ_m \)
\( = 39.05 / 390.5 = 0.1 \)

Question. The conductivity of a \( 0.01 \text{ M} \) solution of acetic acid at 298 K is \( 1.65 \times 10^{-4} \text{ S cm}^{-1} \). Calculate molar conductivity \( (\Lambda_m) \) of the solution.
Answer: \( \Lambda_m = \frac{1000 \kappa}{C} \)
\( \Lambda_m = \frac{1.65 \times 10^{-4} \text{ S cm}^{-1} \times 1000 \text{ cm}^3 \text{ L}^{-1}}{0.01 \text{ mol L}^{-1}} \)
\( = 16.5 \text{ S cm}^2 \text{ mol}^{-1} \)

Question. The conductivity of \( 10^{-3} \text{ mol/L} \) acetic acid at \( 25^\circ\text{C} \) is \( 4.1 \times 10^{-5} \text{ S cm}^{-1} \). Calculate its degree of dissociation if \( \Lambda^\circ_m \) for acetic acid at \( 25^\circ\text{C} \) is \( 390.5 \text{ S cm}^2 \text{ mol}^{-1} \).
Answer: \( \Lambda_m = \frac{1000 \kappa}{C} \)
\( \Lambda_m = \frac{1000 \text{ cm}^3 \text{ L}^{-1} \times 4.1 \times 10^{-5} \text{ S cm}^{-1}}{10^{-3} \text{ mol L}^{-1}} \)
\( = 41 \text{ S cm}^2 \text{ mol}^{-1} \)
\( \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} \)
\( \alpha = \frac{41}{390.5} = 0.105 \)

Question. Why on dilution the \( \Lambda_m \) of \( CH_3COOH \) increases drastically, while that of \( CH_3COONa \) increases gradually?
Answer: In case of \( CH_3COOH \) which is a weak electrolyte, the number of ions increases on dilution due to an increase in degree of dissociation resulting in drastic increase in \( \Lambda_m \).
\[ CH_3COOH + H_2O \rightarrow CH_3COO^- + H_3O^+ \]
In the case of \( CH_3COONa \) which is a strong electrolyte, the number of ions remains the same but the inter-ionic attraction decreases resulting in gradual increase in \( \Lambda_m \).

Question. Solutions of two electrolytes 'A' and 'B' are diluted. The limiting molar conductivity of 'B' increases 1.5 times while that of 'A' increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Answer: 'B' is a strong electrolyte.
B is a strong electrolyte which is completely dissociated into ions, but on dilution interionic forces overcome and ions are free to move. So there is slight increase in molar conductivity on dilution.

Question. The resistivity of a 0.8M solution of electrolyte is \( 5 \times 10^{-3} \Omega\text{cm} \). Calculate its molar conductivity.
Answer: Resistivity \( (\rho) = 5 \times 10^{-3} \Omega\text{ cm} \)
Conductivity of solution (\( \kappa \))
\( = \frac{1}{\text{Resistivity}} \)
\( = \frac{1}{5 \times 10^{-3} \Omega\text{ cm}} \)
\( = 0.2 \times 10^3 \Omega^{-1}\text{cm}^{-1} \)
\( = 200 \Omega^{-1}\text{cm}^{-1} \)
Molar conductivity
\( \Lambda_m = \frac{1000 \times \kappa}{M} \)
\( = \frac{1000 \times 200}{0.8} = 2.5 \times 10^5 \Omega^{-1}\text{cm}^2\text{mol}^{-1} \)

Question. In a galvanic cell, the following cell reaction occurs:
\( Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s) \)
\( E^\circ_{\text{cell}} = +1.56 \text{ V} \)
(i) Is the direction of flow of electrons from zinc to silver or silver to zinc?
(ii) How will concentration of \( Zn^{2+} \) ions and \( Ag^+ \) ions be affected when the cell functions?
Answer: (i) Zinc to silver
(ii) Concentration of \( Zn^{2+} \) ions will increase and \( Ag^+ \) ions will decrease.

Question. Calculate the emf of the following cell at 298 K
\( Cr(s) | Cr^{3+} (0.1\text{M}) || Fe^{2+} (0.01\text{M}) | Fe(s) \)
[Given: \( E^\circ_{\text{cell}} = + 0.30 \text{ V} \)]
Answer: \( 2Cr(s) + 3Fe^{2+}(aq) \rightarrow 3Fe(s) + 2Cr^{3+}(aq) \)
Where, \( n = 6 \)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303RT}{nF} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} \)
\( E_{\text{cell}} = 0.30 \text{ V} - \frac{0.059}{6} \text{ V} \log \frac{[10^{-1}]^2}{[10^{-2}]^3} \)
\( E_{\text{cell}} = 0.26 \text{ V} \)

Question. Explain redox potential.
Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

Answer: Redox potential (also known as reduction / oxidation potential) is a measure of the tendency of a chemical species to acquire electrons from or lose electrons to an electrode and thereby be reduced or oxidised respectively. Redox potential is measured in volts (V), or millivolts (mV). The more positive the reduction potential of a species, the greater the species' affinity for electrons and tendency to be reduced.
The higher the reduction potential, the higher is its tendency to get reduced. Hence, the order of oxidising power is:
\( BrO_4^- > IO_4^- > ClO_4^- \)

Question. Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes:
\( Ag^+(aq) + e^- \rightarrow Ag(s); \quad E^\circ = 0.80 \text{ V} \)
\( H^+(aq) + e^- \rightarrow \frac{1}{2} H_2(g); \quad E^\circ = 0.00 \text{ V} \)
On the basis of their standard electrode potential values, which reaction is feasible at cathode and why?
Answer: \( Ag^+(aq) + e^- \rightarrow Ag(s) \)
Because it has higher reduction potential.
Detailed Answer: As reaction with higher value of standard electrode potential occurs at cathode, Ag gets reduced. So, the reaction occurring at cathode is
\( Ag^+(aq) + e^- \rightarrow Ag(s) \)

Question. Calculate \( E^\circ_{\text{cell}} \) for the following reaction at 298 K:
\( 2Cr(s) + 3Fe^{2+}(0.01\text{M}) \rightarrow 2Cr^{3+}(0.01\text{M}) + 3Fe(s) \)
Given: \( E_{\text{cell}} = 0.261 \text{ V} \)
Answer: Nernst Equation:
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Prod.}]}{[\text{React.}]} \)
\( E^\circ_{\text{cell}} = E_{\text{cell}} + \frac{0.059}{n} \log \frac{[\text{Prod.}]}{[\text{React.}]} \)
\( = 0.261 + \frac{0.059}{6} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} \)
\( = 0.261 + \frac{0.059}{6} \log \frac{(10^{-2})^2}{(10^{-2})^3} \)
\( = 0.261 + \frac{0.059}{6} \log 10^2 \)
\( = 0.261 + \frac{0.059 \times 2}{6} \)
\( = 0.261 + 0.01966 \)
\( = 0.28068\text{V} \approx 0.281\text{V} \)

Question. Iron displaces copper from copper sulphate solution but Pt does not why?
Answer: Electrode potential of Fe is more than electrode potential of Cu. So, Fe displaces Cu from copper sulphate while electrode potential of Pt is less than Cu. Due to this reason, Pt cannot displace Cu from copper sulphate.

Question. Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell.
Answer: Mercury cell
Anode: \( Zn(Hg) + 2OH^- \rightarrow ZnO(s) + H_2O + 2e^- \)
Cathode: \( HgO + H_2O + 2e^- \rightarrow Hg(l) + 2OH^- \)
Detailed Answer:
Mercury cell is generally used in hearing aids.
At anode: \( Zn(Hg) + 2OH^- \rightarrow ZnO(s) + H_2O + 2e^- \)
At cathode: \( HgO(s) + H_2O + 2e^- \rightarrow Hg(l) + 2OH^- \)
Overall reaction: \( Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l) \)

Question. Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell.
Answer: Dry cell/Leclanche cell
Anode: \( Zn(s) \rightarrow Zn^{2+} + 2e^- \)
Cathode: \( MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3 \)
Detailed Answer:
The cell which is used in the transistors is Dry cell.
At anode: \( Zn(s) \rightarrow Zn^{2+} + 2e^- \)
At cathode: \( MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3 \)
Ammonia produced in the reaction forms a complex with \( Zn^{2+} \) ion.
\( Zn^{2+} + 4NH_3 \rightarrow [Zn(NH_3)_4]^{2+} \)

Question. From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell.
Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life?
Answer:
(i) Mercury cell
(ii) Fuel cell
(iii) Lead storage cell
(iv) Dry cell

Question. Write the electrode reactions for a \( H_2-O_2 \) fuel cell.
Answer:
At anode,
\( 2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^- \)
At cathode,
\( O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq) \)
Net cell reaction,
\( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \)

Short Answer Type Questions-II 

Question. (i) Give Debye Huckel Onsager equation for strong electrolyte.
(ii) Given are the conductivity and molar conductivity of NaCl solutions at 298K at different concentrations:

Compare the variation of conductivity and molar conductivity of NaCl solutions on dilution. Give reason.
Answer: (i) Debye Huckel Onsager equation for strong electrolyte is:
\( \Lambda_m = \Lambda^\circ_m - b\sqrt{C} \)
Where \( \Lambda_m \) = Molar conductivity
\( \Lambda^\circ_m \) = Molar conductivity at infinite dilution
\( b \) = Constant
\( C \) = Concentration of solution
(ii) Conductivity of NaCl decreases on dilution as the number of ions per unit volume decreases. Whereas molar conductivity of NaCl increases on dilution as on dilution the interionic interactions overcome and ions are free to move.

Question. The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is \( 4.55 \times 10^3 \) ohm. Calculate its molar conductivity.
Answer: \( A = \pi r^2 \)
\( = 3.14 \times 0.5 \times 0.5 \text{ cm}^2 \)
\( = 0.785 \text{ cm}^2 \)
\( l = 45.5 \text{ cm} \)
\( G^* = l/A = 45.5 \text{ cm}/0.785 \text{ cm}^2 \)
\( = 57.96 \text{ cm}^{-1} \)
\( \kappa = G^*/R \)
\( = 57.96 \text{ cm}^{-1} / 4.55 \times 10^3 \Omega \)
\( = 1.27 \times 10^{-2} \text{ S cm}^{-1} \)
\( \Lambda_m = \kappa \times 1000/C \)
\( = [1.27 \times 10^{-2} \text{ S cm}^{-1}] \times 1000 / 0.05 \text{ mol/cm}^3 \)
\( = 254.77 \text{ S cm}^2 \text{ mol}^{-1} \)

Question. (i) State the law which helps to determine the limiting molar conductivity of weak electrolyte.
(ii) Calculate limiting molar conductivity of \( CaSO_4 \) (limiting molar conductivity of calcium and sulphate ions are 119.0 and 160.0 \( \text{S cm}^2\text{ mol}^{-1} \) respectively)
Answer: Kohlrausch law of independent migration of ions:
(i) The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contribution of the anions and cations of the electrolyte.
(ii) \( \Lambda^\circ_{m (CaSO_4)} = \Lambda^\circ_{Ca^{2+}} + \Lambda^\circ_{SO_4^{2-}} \)
\( = 119.0 \text{ S cm}^2\text{ mol}^{-1} + 160.0 \text{ S cm}^2\text{ mol}^{-1} \)
\( = 279.0 \text{ S cm}^2\text{ mol}^{-1} \)

Question. Consider the following reaction:
\( Cu(s) + 2Ag^+(aq) \rightarrow 2Ag(s) + Cu^{2+}(aq) \)
(i) Depict the galvanic cell in which the given reaction takes place.
(ii) Give the direction of flow of current.
(iii) Write the half-cell reactions taking place at cathode and anode.
Answer:
(i) \( Cu(s) | Cu^{2+}(aq) || Ag^+(aq) | Ag(s) \)
(ii) Current will flow from silver to copper electrode in the external circuit.
(iii) Cathode: \( 2Ag^+(aq) + 2e^- \rightarrow 2Ag(s) \)
Anode: \( Cu(s) \rightarrow Cu^{2+}(aq) + 2e^- \)

Question. (a) The cell in which the following reaction occurs:
\( 2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s) \)
has \( E^\circ_{\text{cell}} = 0.236 \text{ V} \) at 298 K. Calculate the standard Gibb's energy of the cell reaction.
(Given: \( 1\text{F} = 96,500 \text{ C mol}^{-1} \))
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours?
(Given: \( 1\text{F} = 96,500 \text{ C mol}^{-1} \))
Answer:
(a) \( \Delta G^\circ = -nFE^\circ_{\text{cell}} \)
\( n = 2 \)
\( \Delta G^\circ = -2 \times 96500 \text{ C /mol} \times 0.236 \text{ V} \)
\( = -45548 \text{ J/mol} \)
\( = -45.548 \text{ kJ/mol} \)
(b) \( Q = I \times t = 0.5 \times 2 \times 60 \times 60 \)
\( = 3600 \text{ C} \)
\( 96500 \text{ C} = 6.023 \times 10^{23} \text{ electrons} \)
\( 3600 \text{ C} = 2.25 \times 10^{22} \text{ electrons} \)
Detailed Answer:
(a) \( 2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+} \)
\( 2I^- \rightarrow I_2 + 2e^- \)
For the given cell reaction, \( n = 2 \).
\( \Delta G^\circ = -nFE^\circ_{\text{cell}} \)
\( = -2 \times 96500 \times 0.236 \)
\( = -45548 \text{ J mol}^{-1} \)
\( = -45.55 \text{ kJ mol}^{-1} \)
(b) \( I = 0.5\text{A} \)
\( t = 2 \text{ hours} = 2 \times 60 \times 60 \text{ s} = 7200\text{s} \)
\( Q = It \)
\( = 0.5 \times 7200 \)
\( = 3600 \text{ coulombs} \)
A flow of 96500 C is equal to flow of 1 mole of electrons is \( 6.023 \times 10^{23} \) electrons.
\( \therefore 3600 \text{ C} \) is equivalent to electrons
\( = \frac{6.023 \times 10^{23}}{96500} \times 3600 \)
\( = 2.246 \times 10^{22} \text{ electrons} \)

Question. Calculate \( \Delta_r G^\circ \) and \( \log K_c \) for the following reaction at 298 K.
\( 2Cr(s) + 3Fe^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Fe(s) \)
[\( E^\circ_{\text{cell}} = 0.30 \text{ V} \), \( 1\text{F} = 96500 \text{ C mol}^{-1} \)]
Answer: \( \Delta_r G^\circ = -nFE^\circ_{\text{cell}} \), \( n = 6 \)
\( = -6 \times 96500 \text{ C/mol} \times 0.30 \text{ V} \)
\( = -173700 \text{ J/mol} = -173.7 \text{ kJ/mol} \)
\( E^\circ_{\text{cell}} = 0.059\text{V}/n \times \log K_c \)
\( \log K_c = 0.30 \text{ V} \times 6 / 0.059 \text{ V} = 30.5 \)

Question. Calculate the emf of the following cell at \( 25^\circ \text{C} \):
\( Fe | Fe^{2+} (0.001 \text{ M}) || H^+ (0.01 \text{ M}) | H_2(g) (1\text{bar}) | Pt(s) \)
\( E^\circ (Fe^{2+} / Fe) = -0.44 \text{ V} \); \( E^\circ (H^+ / H_2) = 0.00 \text{ V} \)
Answer: Cell reaction is
\( Fe(s) + 2H^+(aq) \rightarrow Fe^{2+}(aq) + H_2(g) \)
\( E^\circ_{\text{cell}} = 0.00 - (-0.44) = 0.44 \text{ V} \)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \frac{[Fe^{2+}]}{[H^+]^2} \)
\( = 0.44 \text{ V} - \frac{0.0591}{2} \text{ V} \log \frac{0.001}{(0.01)^2} \)
\( = 0.44 \text{ V} - 0.02955 \text{ V} \)
\( E_{\text{cell}} = 0.41045 \text{ V} \)

Question. A galvanic cell consists of a metallic zinc plate immersed in 0.1 M \( Zn(NO_3)_2 \) solution and metallic plate of lead in 0.02 M \( Pb(NO_3)_2 \) solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell.
(Given: \( E^\circ_{Zn^{2+}/Zn} = -0.76 \text{ V} \); \( E^\circ_{Pb^{2+}/Pb} = -0.13 \text{ V} \))
Answer: Anode reaction: \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \)
Cathode reaction: \( Pb^{2+}(aq) + 2e^- \rightarrow Pb(s) \)
Cell representation:
\( Zn(s) | Zn^{2+}(aq) || Pb^{2+}(aq) | Pb(s) \)
According to Nernst equation:
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Pb^{2+}]} \)
\( E_{\text{cell}} = [-0.13 - (-0.76)] - \frac{0.0591}{2} \log \frac{0.1}{0.02} \)
\( = 0.63 - 0.02955 \times \log 5 \)
\( = 0.63 - 0.02955 \times 0.6990 \)
\( = 0.63 - 0.0206 = 0.6094 \text{ V} \)

Question. When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes \( ZnSO_4 \) and \( CuSO_4 \) connected in series, 2 g of Cu were deposited at the cathode of cell B. How long did the current flow? What mass of Zn was deposited at cathode of cell A?
[Atomic mass: Cu = 63.5 g mol\(^{-1}\), Zn = 65 g mol\(^{-1}\); \( 1\text{F} = 96500 \text{ C mol}^{-1} \)]
Answer: \( Zn^{2+}(aq) + 2e^- \rightarrow Zn(s) \) (1 mol)
\( Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \) (1 mol) (2 gm given)
The charge Q on a mole of electrons, \( Q = nF \)
Calculation of time for the flow of current:
\( n = 1 \text{ mol} \)
\( Q = 1 \times 96500 \text{ C mol}^{-1} = 96500 \text{ C} \)
Molar mass of Cu = 63.5 gm mol\(^{-1}\)
\( \because \) 63.5 gm of Cu is deposited by electric charge = 96500C
\( \therefore \) 2 gm of Cu is deposited by electric charge
\( = \frac{96500}{63.5} \times 2 = 3039.37 \text{ C} \)
Let 2 A of current be passed for time \( t \), quantity of electricity used = \( 2\text{A} \times t = 3039.37 \text{ C} \)
or, \( t = \frac{3039.37\text{C}}{2} = 1519.68 \text{ s} \)
\( = 25 \text{ min } 33 \text{ s} \)
Calculation of mass of Zn deposited:
\( \frac{W_1}{W_2} = \frac{E_1}{E_2} = \frac{\text{Equivalent weight of Zn}}{\text{Equivalent weight of Cu}} \)
\( = \frac{\text{Molar mass of Zn} / \text{Charge on Zn}}{\text{Molar mass of Cu} / \text{Charge on Cu}} \)
Amount of Zn deposited:
\( = 2 \times \frac{\frac{65}{2}}{\frac{63.5}{2}} = 2.0472 \text{ g} \)

Question. (a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of \( AgNO_3 \) for 15 minutes.
(Given: Molar mass of Ag = 108 g mol\(^{-1}\), \( 1\text{F} = 96500 \text{ C mol}^{-1} \))
(b) Define fuel cell.
Answer: (a) \( m = ZIt \)
\( = \frac{108 \times 2 \times 15 \times 60}{1 \times 96500} \)
\( = 2.01 \text{ g} \) (or any other correct method)
(b) Cells that convert the energy of combustion of fuels directly into electrical energy.
Detailed Answer:
(a) \( t = 900 \text{ s} \)
Charge = Current \( \times \) Time = \( 2 \times 900 = 1800 \text{ C} \)
According to the reaction
\( Ag^+(aq) + e^- \rightarrow Ag(s) \)
We require 1 F to deposit 1 mol or 108 g of Ag
For 1800 C, the mass of Ag deposited will be \( = \frac{108 \times 1800}{1 \times 96500} = 2.0145 \text{ g} \)
(b) Fuel cell is the name given to the galvanic cells which are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy.

Question. The electrolysis of a metal salt solution was carried out by passing a current of 4 A for 45 minutes. It resulted in deposition of 2.977 g of a metal. If atomic mass of the metal in 106.4 g mol\(^{-1}\), calculate the charge on the metal cation.
Answer: Let the charge on the metal ion = \( n^+ \)
Reduction half-reaction,
\( M^{n+} + ne^- \rightarrow M \)
(1mol) (n mol) (106.4 g)
Quantity of electricity required for depositing 106.4 g of metal = \( n \times 96500 \text{ C} \)
Quantity of electricity required for depositing 2.977 g of metal = \( \frac{n \times 96500 \times 2.977}{106.4} = n \times 2700 \)
Quantity of electricity passed
\( = 4 \times 45 \times 60 \text{ C} \)
\( = 10800 \text{ C} \)
Applying law of conservation of charge,
\( 10800 = n \times 2700 \)
\( n = \frac{10800}{2700} = 4 \)
Charge on metal ion = +4

Long Answer Type Questions 

Question. (a) The electrical resistance of a column of 0.05 M KOH solution of length 50 cm and area of cross-section 0.625 cm\(^2\) is \( 5 \times 10^3 \) ohm. Calculate its resistivity, conductivity and molar conductivity.
(b) Predict the products of electrolysis of an aqueous solution of \( CuCl_2 \) with platinum electrodes.
(Given: \( E^\circ_{Cu^{2+}/Cu} = + 0.34 \text{ V} \), \( E^\circ_{(\frac{1}{2}Cl_2/Cl^-)} = + 1.36 \text{ V} \); \( E^\circ_{H^+/H_2(g),Pt} = 0.00 \text{ V} \), \( E^\circ_{(\frac{1}{2}O_2/H_2O)} = + 1.23 \text{ V} \))
Answer: (a) Given: A = 0.625 cm\(^2\), l = 50 cm
R = \( 5 \times 10^3 \text{ ohm} \), \( \rho \) = ?
m = 0.05 m, \( \kappa \) = ?
\( \Lambda_m \) = ?
Cell constant = \( \frac{l}{A} = \frac{50 \text{ cm}}{0.625 \text{ cm}^2} = 80 \text{ cm}^{-1} \)
Resistivity = \( \frac{\text{R}}{\text{cell constant}} \) or \( \frac{R \times A}{l} \)
\( \implies \frac{5 \times 10^3 \text{ ohm}}{80 \text{ cm}^{-1}} \)

\( \implies \) 62.5 ohm cm.
Conductivity = \( \frac{1}{\text{Resistance}} \times \frac{l}{A} \)
\( = \frac{1}{5 \times 10^3} \times 80 = 0.016 \text{ S cm}^{-1} \)
Molar conductivity (\( \Lambda_m \)) = \( \frac{10^3 \kappa}{M} \)
\( \Lambda_m = \frac{\kappa}{M} \times 1000 = \frac{10 \times 1000}{625 \times 0.05} = 320 \text{ S cm}^2 \text{ mol}^{-1} \)
(b) Given: \( E^\circ_{Cu^{2+}/Cu} = +0.34 \text{ V} \)
\( E^\circ_{(\frac{1}{2}Cl_2/Cl^-)} = +1.36 \text{ V} \)
\( E^\circ_{H^+/H_2(g)}, Pt = 0.00 \text{ V} \), \( E^\circ_{(O_2/H_2O)} = +1.23 \text{ V} \)
At cathode:
\( Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu(s); E^\circ = 0.34 \text{ V} \)
\( H^+_{(aq)} + e^- \rightarrow \frac{1}{2}H_2(g); E^\circ = 0 \text{ V} \)
The reaction with a higher value of E° takes place at the cathode, so deposition of copper will take place at the cathode.
At anode: The oxidation reactions are possible at the anode.
\( Cl^-_{(aq)} \rightarrow \frac{1}{2}Cl_2(g) + e^-; E^\circ = 1.36 \text{ V} \)
\( 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-; E^\circ = +1.23 \text{ V} \)
At the anode the reaction with a lower value of E° is preferred. But due to the over potential of oxygen, Cl\(^-\) gets oxidised at anode to produce Cl\(_2\) gas.

Question. The molar conductivities of acetic acid at 298 K at the concentrations of 0.1 M and 0.001 M are 5.20 and 49.2 S cm\(^2\)mol\(^{-1}\) respectively. Calculate the degree of dissociation of acetic acid at these concentration. Given \( \lambda^\infty_{(H^+)} \) and \( \lambda^\infty_{(CH_3COO^-)} \) are 349.8 and 40.9 ohm\(^{-1}\)cm\(^2\)mol\(^{-1}\) respectively.
Answer: Given
\( \lambda^\infty_{(H^+)} = 349.8 \text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1} \)
\( \lambda^\infty_{(CH_3COO^-)} = 40.9 \text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1} \)
\( \lambda^\infty_{m(CH_3COOH)} = \lambda^\infty_{(CH_3COO^-)} + \lambda^\infty_{(H^+)} \)
\( = 40.9 + 349.8 \)
\( = 390.7 \text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1} \)
At C = 0.1 M
Degree of dissociation
\( \alpha = \frac{\Lambda_m}{\Lambda^\infty_m} \)
\( = \frac{5.20}{390.7} = 0.013 \text{ i.e. 1.3\%} \)
At C = 0.001 M
\( \alpha = \frac{\Lambda_m}{\Lambda^\infty_m} \)
\( = \frac{49.2}{390.7} = 0.125 \text{ i.e. 12.5\%} \)

Question. The resistance of 0.01 M acetic solution when measured in a conductivity cell of cell constant 0.366 cm\(^{-1}\), is found to be 2220 \(\Omega\). Calculate degree of dissociation of acetic acid at this concentration. Also find the dissociation constant of acetic acid. Given that value of \( \lambda^\infty_{H^+} \) and \( \lambda^\infty_{CH_3COO^-} \) as 349.1 and 40.9 \(\Omega^{-1}\)cm\(^2\)mol\(^{-1}\) respectively.
Answer: Conductivity (\( \kappa \)) of 0.01 M acetic acid
\( \kappa = \frac{1}{\text{R}} \times \text{(cell constant)} \)
\( = \frac{1}{2220 \Omega} \times 0.366 \text{ cm}^{-1} \)
\( = 1.648 \times 10^{-4} \Omega^{-1}\text{cm}^{-1} \)
Molar conductivity
\( \Lambda_m = \frac{\kappa \times 1000}{M} \)
\( = \frac{1.648 \times 10^{-4} \times 1000}{0.01} \)
\( = 16.48 \Omega^{-1}\text{cm}^2\text{mol}^{-1} \)
Molar conductivity at infinite dilution
\( \Lambda^\infty_{m(CH_3COOH)} = \lambda^\infty_{H^+} + \lambda^\infty_{CH_3COO^-} \)
\( = 349.1 + 40.9 \)
\( = 390 \Omega^{-1}\text{cm}^2\text{mol}^{-1} \)
Degree of dissociation of acetic acid
\( \alpha = \frac{\Lambda_m}{\Lambda^\infty_m} \)
\( = \frac{16.48}{390} = 0.0422 \)
\( CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \)
Initial conc. (mol/L): \( C \quad 0 \quad 0 \)
Equilibrium conc. (mol/L): \( C - C\alpha \quad C\alpha \quad C\alpha \)
Dissociation constant
\( K = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \)
\( = \frac{C\alpha \times C\alpha}{C - C\alpha} \)
\( = \frac{C\alpha^2}{(1 - \alpha)} \)
\( = \frac{0.01 \times (0.0422)^2}{1 - 0.0422} \)
\( = 1.86 \times 10^{-5} \)

Question. (a) Calculate e.m.f. of the following cell:
\( Zn(s) / Zn^{2+} (0.1 \text{ M}) || (0.01 \text{ M}) Ag^+ / Ag(s) \)
Given: \( E^\circ_{Zn^{2+}/Zn} = -0.76 \text{ V} \), \( E^\circ_{Ag^+/Ag} = + 0.80 \text{ V} \)
[Given: \( \log 10 = 1 \)]
(b) X and Y are two electrolytes. On dilution molar conductivity of 'X' increases 2.5 times while that Y increases 25 times. Which of the two is a weak electrolyte and why?
Answer: (a) \( Zn(s) / Zn^{2+} (0.1 \text{ M}) || (0.01\text{M}) Ag^+ / Ag(s) \)
\( E^\circ_{Zn^{2+}/Zn} = -0.76\text{V} \)
\( E^\circ_{Ag^+/Ag} = + 0.80 \text{ V} \quad \text{emf} = ? \)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Anode}]}{[\text{Cathode}]} \)
\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \)
\( = E^\circ_{Ag/Ag} - E^\circ_{Zn^{2+}/Zn} \)
\( = 0.80 - (-0.76) = 1.56 \text{ V} \)
\( E_{\text{cell}} = 1.56 - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Ag^+]^2} \)
\( = 1.56 - \frac{0.0591}{2} \log \frac{[0.1]}{[0.01]^2} \)
\( = 1.56 - 0.0295 \log 1000 \)
\( = 1.56 - 3(0.0295) = 1.56 - 0.09 = 1.4715 \text{ V} \)
(b) Y is a weak electrolyte as on dilution complete dissociation of weak electrolyte takes place and thus a sharp increase in molar conductivity while in strong electrolyte it has already dissociated completely. So on dilution molar conductivity does not rises much.

Question. \( E^\circ_{\text{cell}} \) for the given redox reaction is 2.71 V.
\( Mg + Cu^{2+}_{(0.01 \text{ M})} \rightarrow Mg^{2+}_{(0.001 \text{ M})} + Cu \)
Calculate \( E_{\text{cell}} \) for the reaction. Write the direction of flow of current when an external opposite potential applied is:
(i) Less than 2.71 V
(ii) Greater than 2.71 V
Answer: \( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log K_c \)
\( = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{10^{-3}}{10^{-2}} = 2.71 + 0.0295 \)
\( E_{\text{cell}} = 2.7395 \text{ V} \)
(i) Cu to Mg / Cathode to anode / Same direction
(ii) Mg to Cu / Anode to cathode / Opposite direction
Detailed Answer:
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left[ \frac{Mg^{2+}}{Cu^{2+}} \right] \)
\( E_{\text{cell}} = 2.71 - \frac{0.059}{2} \log \left[ \frac{0.001}{0.01} \right] \)
\( E_{\text{cell}} = 2.71 - (-0.0295) = 2.74 \text{ V} \)
(i) When external opposite applied voltage is less than 2.71, it is less than \( E^\circ_{\text{cell}} \), therefore, the electrons will flow from the anode to the cathode, and current will flow from cathode (copper electrode) to anode (magnesium electrode).
(ii) When external opposite applied potential is greater than 2.71, which is greater than \( E^\circ_{\text{cell}} \), therefore, the reaction will be reversed, and the current will flow from anode to cathode.

Question. (a) Represent the cell in which the following reaction takes place:
\( 2Al(s) + 3Ni^{2+}(0.1\text{M}) \rightarrow 2Al^{3+}(0.01\text{M}) + 3Ni(s) \)
Calculate its emf if \( E^\circ_{\text{cell}} = 1.41\text{V} \).
(b) How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity \( (\Lambda^\circ_m) \) for weak electrolyte?
Answer: (a) \( Al(s) | Al^{3+}(0.01\text{M}) || Ni^{2+}(0.1\text{ M}) | Ni(s) \)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{6} \log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3} \)
\( E_{\text{cell}} = 1.41\text{V} - \frac{0.059}{6} \log \frac{[0.01]^2}{[0.1]^3} \)
\( E_{\text{cell}} = 1.4198\text{ V} \) or \( E_{\text{cell}} = 1.42\text{ V} \)
(b) \( \Lambda_m \) decreases with increase in concentration for both strong & weak electrolyte. \( \Lambda^\circ_m \) can be obtained for weak electrolyte by applying Kohlrausch law
\( \Lambda^\circ_m = V_+ \lambda^\circ_+ + V_- \lambda^\circ_- \)
Detailed Answer:
(a) \( 2Al(s) + 3Ni^{2+}(0.1\text{M}) \rightarrow 2Al^{3+}(0.01\text{M}) + 3Ni(s) \)
\( Al(s) | Al^{3+} || Ni^{2+} | Ni(s) \rightarrow \) Cell reaction
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{6} \log \frac{[0.01]^2}{[0.1]^3} \)
\( = 1.41 - \frac{0.0591}{6} \log 0.1 \)
\( = 1.41 + \frac{0.0591}{6} \log 10 \)
\( = 1.41 + \frac{0.0591}{6} \)
\( = 1.41 + 0.00985 = 1.42 \text{ V} \)
(b) When the concentration of weak electrolyte becomes very low, its degree of ionization rises. This increase leads to increase in the number of ions in the solution. Thus, the molar conductivity rises sharply of a weak electrolyte at low concentration. The molar conductivity of strong electrolyte decreases a bit with an increase in concentration. This is due to increase in interionic attraction due to higher number of ions per unit volume. On dilution, ions move apart, weakening interionic attractions and thus conductance increases.
Limiting molar conductivity for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions.

Question. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:
\( Sn(s) | Sn^{2+} (0.004 \text{ M}) || H^+ (0.020 \text{ M}) | H_2(g) (1 \text{ bar}) | Pt(s) \)
(Given: \( E^\circ_{Sn^{2+}/Sn} = - 0.14\text{V} \))
(b) Give reasons:
(i) On the basis of E° values, \( O_2 \) gas should be liberated at anode but it is \( Cl_2 \) gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of \( CH_3COOH \) decreases on dilution.
Answer: (a) \( Sn + 2H^+ \rightarrow Sn^{2+} + H_2 \) (Equation must be balanced)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{[Sn^{2+}]}{[H^+]^2} \)
\( E_{\text{cell}} = [0 - (- 0.14)] - 0.0295 \log \frac{(0.004)}{(0.02)^2} \)
\( = 0.14 - 0.0295 \log 10 = 0.11 \text{ V} / 0.1105 \text{ V} \)
(b) (i) Due to over potential/ over voltage of \( O_2 \)
(ii) The number of ions per unit volume decreases.
Detailed Answer:
(a) \( Sn(s) | Sn^{2+} (0.004 \text{ M}) || H^+ (0.020 \text{ M}) | H_2(g)(1 \text{ bar}) | Pt(s) \)
\( E^\circ_{\text{cell}} = E^\circ_{(H^+/H_2)} - E^\circ_{(Sn^{2+}/Sn)} \)
\( = 0.00 - (- 0.14) \)
\( = +0.14\text{V} \)
\( Sn(s) \rightarrow Sn^{2+}(aq) + 2e^- \)
\( 2H^+(aq) + 2e^- \rightarrow H_2(g) \)
\( \overline{Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)} \)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[Sn^{2+}]}{[H^+]^2} \)
\( = 0.14 - \frac{0.0591}{2} \log \frac{(4 \times 10^{-3})}{(2 \times 10^{-2})^2} \)
\( = 0.14 - 0.0295 \log 10 \)
\( = 0.14 - 0.0295 \)
\( = 0.1105 \text{ V} \)
(b) (i) \( NaCl \rightarrow Na^+ + Cl^- \)
\( H_2O \rightleftharpoons H^+ + OH^- \)
The value of E° of \( O_2 \) is higher than \( Cl_2 \) but \( O_2 \) is evolved from \( H_2O \) only when the higher voltage is applied. So, because of this \( Cl_2 \) is evolved instead of \( O_2 \).
(ii) Conductivity varies with the change in the concentration of the electrolyte. The number of ions per unit volume decreases on dilution. So, conductivity decreases with decrease in concentration. Therefore, conductivity of \( CH_3COOH \) decreases on dilution.

Question. (a) Calculate \( E^\circ_{\text{cell}} \) for the following reaction at 298K:
\( 2Al(s) + 3Cu^{2+} (0.01\text{M}) \rightarrow 2Al^{3+} (0.01\text{M}) + 3Cu(s) \)
Given: \( E^\circ_{\text{cell}} = 1.98 \text{ V} \)
(b) Using the E° values of A and B, predict which is better for coating the surface of iron [\( E^\circ_{(Fe^{2+}/Fe)} = -0.44\text{V} \)] to prevent corrosion and why?
Given: \( E^\circ_{(A^{2+}/A)} = -2.37\text{V} \): \( E^\circ_{(B^{2+}/B)} = -0.14\text{V} \)
Answer: (a) \( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3} \)
\( E_{\text{cell}} = E^\circ_{\text{cell}} + \frac{0.0591}{n} \log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3} \)
\( E_{\text{cell}} = 1.98 \text{ V} + \frac{0.0591}{6} \log \frac{(0.01)^2}{(0.01)^3} \)
\( E_{\text{cell}} = 1.98 \text{ V} + \frac{0.0591}{6} \log 10^2 \)
\( E_{\text{cell}} = 1.98 \text{ V} + \frac{0.0591}{6} \times 2 \times \log 10 \)
[\( \because \log 10 = 1 \)]
\( E_{\text{cell}} = 1.98 \text{ V} + \frac{0.0591\text{ V}}{6} \times 2 \)
\( E_{\text{cell}} = 1.98 \text{ V} + 0.0197 \text{ V} \)
\( E_{\text{cell}} = 1.9997 \text{ V} \)
(b) A, because its \( E^\circ \) value is more negative

Question. (a) The conductivity of 0.001 mol L\(^{-1}\) solution of \( CH_3COOH \) is \( 3.905 \times 10^{-5} \text{ S cm}^{-1} \). Calculate its molar conductivity and degree of dissociation (\( \alpha \)).
Given \( \lambda^\circ (H^+) = 349.6 \text{ S cm}^2 \text{ mol}^{-1} \) and \( \lambda^\circ(CH_3COO^-) = 40.9 \text{ S cm}^2 \text{ mol}^{-1} \)
(b) Define electrochemical cell. What happens if external potential applied becomes greater than \( E^\circ_{\text{cell}} \) of electrochemical cell?
Answer: (a) \( \Lambda_m = \kappa \times 1000/C \)
\( = \frac{3.905 \times 10^{-5} \times 1000}{0.001} \)
\( = 39.05 \text{ S cm}^2/\text{mol} \)
\( CH_3COOH \rightarrow CH_3COO^- + H^+ \)
\( \Lambda^\circ_{CH_3COOH} = \lambda^\circ_{CH_3COO^-} + \lambda^\circ_{H^+} \)
\( = 40.9 + 349.6 \)
\( \Lambda^\circ_{CH_3COOH} = 390.5 \text{ S cm}^2/\text{mol} \)
\( \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{39.05}{390.5} = 0.1 \)
(b) Device used for the production of electricity from energy released during spontaneous chemical reaction and the use of electrical energy to bring about a chemical change.
The reaction gets reversed / It starts acting as an electrolytic cell & vice - versa.

Question. (a) The conductivity of 0.001 mol L\(^{-1}\) acetic acid is \( 4.95 \times 10^{-5} \text{ S cm}^{-1} \). Calculate the dissociation constant if \( \Lambda^\circ_m \) for acetic acid is \( 390.5 \text{ S cm}^2 \text{ mol}^{-1} \).
(b) Write Nernst equation for the reaction at \( 25^\circ\text{C} \):
\( 2Al(s) + 3Cu^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Cu(s) \)
(c) What are secondary batteries? Give an example.
Answer: (a) \( \Lambda_m = \frac{\kappa}{C} = \frac{4.95 \times 10^{-5}\text{S cm}^{-1}}{0.001\text{ mol L}^{-1}} \times \frac{1000\text{ cm}^3}{\text{L}} \)
\( = 49.5\text{ S cm}^2\text{mol}^{-1} \)
\( \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{49.5\text{ S cm}^2\text{mol}^{-1}}{390.5\text{ S cm}^2\text{mol}^{-1}} = 0.126 \)
\( K = \frac{C\alpha^2}{(1-\alpha)} = \frac{0.001\text{mol L}^{-1} \times (0.126)^2}{1-0.126} \)
\( = 1.8 \times 10^{-5} \text{ mol L}^{-1} \)
(If \( K= C\alpha^2 \), then \( K= 1.6 \times 10^{-5} \text{ mol L}^{-1} \))
(b) \( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{6} \log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3} \)
(c) Batteries which are rechargeable
Example - Lead storage, Ni-Cd batteries (Or any other one example).

Question. (a) For the reaction
\( 2AgCl(s) + H_2(g) (1\text{ atm}) \rightarrow 2Ag(s) + 2H^+(0.1\text{ M}) + 2Cl^-(0.1\text{ M}) \),
\( \Delta G^\circ = -43600 \text{ J} \) at \( 25^\circ\text{C} \).
Calculate the e.m.f. of the cell.
[\( \log 10^{-n} = -n \)]
(b) Define fuel cell and write its two advantages.
Answer: (a) \( \Delta G^\circ = -nFE^\circ \)

\( \implies -43600 = -2 \times 96500 \times E^\circ \)
\( E^\circ = 0.226 \text{ V} \)
\( E = E^\circ - \frac{0.059}{2} \log \frac{([H^+]^2 [Cl^-]^2)}{[H_2]} \)
\( = 0.226 - \frac{0.059}{2} \log \frac{[(0.1)^2 \times (0.1)^2]}{1} \)
\( = 0.226 - \frac{0.059}{2} \log 10^{-4} \)
\( = 0.226 + 0.118 = 0.344 \text{ V} \) (Deduct half mark if unit is wrong or not written)
(b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol etc.) directly into electrical energy are called fuel cells.
Advantages: High efficiency, non polluting (or any other suitable advantage)
Detailed Answer:
(a) \( \Delta G^\circ = -43600 \text{ J} \)
No. of electrons (n) = 2
F = 96500 C
\( \because \Delta G^\circ = -nFE^\circ \)

\( \implies -43600 = -2 \times 96500 \times E^\circ \)
\( \therefore E^\circ = \frac{-43600}{-2 \times 96500} = 0.226 \text{V} \)
For the reaction,
\( 2AgCl(s) + H_2(g) (1 \text{ atm}) \rightarrow 2Ag(s) + 2H^+(0.1\text{M}) + 2Cl^-(0.1\text{M}) \)
From Nernst equation
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{[H^+]^2[Cl^-]^2}{[H_2]} \)
(\( \because \) Concentration of solids are taken as unity)
\( = 0.226 - \frac{0.059}{2} \log \frac{(0.1)^2(0.1)^2}{1} \)
\( = 0.226 - \frac{0.059}{2} \log 10^{-4} \)
\( = 0.226 + 0.0295 \log 10^4 \)
\( = 0.226 + 0.118 \)
\( = 0.344 \text{ V} \)
(b) Fuel cell: The cell which converts chemical energy of a fuel directly into electrical energy is called fuel cell.
Advantages of fuel cell:

• Pollution free working
• High efficiency